 # Reduce each of the following equations to normal form : `
Question:

Reduce each of the following equations to normal form :

(i) $x+y-2=0$

(ii) $\mathrm{x}+\mathrm{y}+\sqrt{2}=0$

(iii) $x+5=0$

(iv) $2 y-3=0$

(v) $4 x+3 y-9=0$

Solution:

$\Rightarrow x+y=2$

If the equation is in the form of ax + by = c, to get into the normal form we should divide

it by $\sqrt{a^{2}+b^{2}}$  so now

Divide by $\sqrt{1^{2}+1^{2}}=\sqrt{2}$

$\Rightarrow \frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}=\frac{2}{\sqrt{2}}$

$\Rightarrow \frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}=\sqrt{2}$

This is in the form of $x \cos \alpha+y \sin \alpha=p$, where $\alpha=\frac{\pi}{4}$ and $p=\sqrt{2}$

Conclusion: $\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}=\sqrt{2}$ is the normal form of $x+y-2=0$

(ii) $\mathrm{x}+\mathrm{y}+\sqrt{2}=0$

$\Rightarrow x+y=-\sqrt{2}$

If the equation is in the form of $a x+b y=c$, to get into the normal form, we should divide it by $\sqrt{a^{2}+b^{2}}$, so now

Divide by $\sqrt{1^{2}+1^{2}}=\sqrt{2}$

Our new equation is $\frac{\mathrm{x}}{-\sqrt{2}}+\frac{\mathrm{y}}{-\sqrt{2}}=1$

This is in the form of $x \cos \alpha+y \sin \alpha=p$, where $\alpha=\frac{5 \pi}{4}$ and $p=1$

Conclusion: $\frac{\mathrm{x}}{-\sqrt{2}}+\frac{\mathrm{y}}{-\sqrt{2}}=1$ is the normal form of $\mathrm{x}+\mathrm{y}+\sqrt{2}=0$

(iii) $\Rightarrow-x=5$

If the equation is in the form of $a x+b y=c$, to get into the normal form, we should divide it by $\sqrt{a^{2}+b^{2}}$, so now

Divide the equation by $\sqrt{1^{2}+0^{2}}=1$

Our new equation is $-x=5$

This is in the form of $x \cos \alpha+y \sin \alpha=p$, where $\alpha=\pi$ and $p=5$

Conclusion: $-x=5$ is the normal form of $x+5=0$

(iv) $\Rightarrow 2 \mathrm{y}=3$

If the equation is in the form of $a x+b y=c$, to get into the normal form, we should divide it by $\sqrt{a^{2}+b^{2}}$, so now

Divide by $\sqrt{2^{2}+0^{2}}=2$

Our new equation is $\mathrm{y}=\frac{3}{2}$

This is in the form of $x \cos \alpha+y \sin \alpha=p$, where $\alpha=\frac{\pi}{2}$ and $p=\frac{3}{2}$

Conclusion: $\mathrm{y}=\frac{3}{2}$ is the normal form of $2 \mathrm{y}=3$

(v) $\Rightarrow 4 \mathrm{x}+3 \mathrm{y}-9=0$

If the equation is in the form of $a x+b y=c$, to get into the normal form, we should divide it by $\sqrt{a^{2}+b^{2}}$, so now

Divide by $\sqrt{4^{2}+3^{2}}=5$

Our new equation is $\frac{4}{5} \mathrm{x}+\frac{3}{5} \mathrm{y}=\frac{9}{5}$

This is in the form of $x \cos \alpha+y \sin \alpha=p$, where

$\alpha=\sin ^{-1}\left(\frac{3}{5}\right)$ or $\alpha=\cos ^{-1}\left(\frac{4}{5}\right)$ and $\mathrm{p}=\frac{9}{5}$

Conclusion: $\frac{4}{5} \mathrm{x}+\frac{3}{5} \mathrm{y}=\frac{9}{5}$ is the normal form of $4 \mathrm{x}+3 \mathrm{y}-9=0$